# Initial speed is zero and so is power?

If I want to accelerate something from standstill to max speed, with a constant force (acceleration and mass don't change), the equation P = F * v would say that in the beginning we use 0 W power. How is that possible? Since power is the rate of transference of energy to the body (J/s), it would appear that that rate should be constant.

Why is it that a fast moving body requires more power to accelerate than a still body? And if we would like to know how much power we need to accelerate that body, should we then use the maximum speed?

• Do you know calculus? – David White May 29 '19 at 16:08

If you want to accelerate a body from $$v$$ to $$v+\Delta v$$, the associated change in energy is

$$\Delta E=E\left(v+\Delta v\right)-E\left(v\right)=\dfrac{1}{2}m\left(v+\Delta v\right)^{2}-\dfrac{1}{2}mv^{2}\approx mv \Delta v$$

You can see that for larger $$v$$ it costs more energy for the same increment $$\Delta v$$. Essentially the power is

$$\dfrac{\Delta E}{\Delta t}=\underbrace{m\dfrac{\Delta v}{\Delta t}}_{ma=F}v=Fv$$

in the beginning we use 0 power. How is that possible?

We apply a force, but, at the very beginning, the power is 0, and is small as long as the speed is small.

Since power is the rate of transference of energy to the body (J/s), it would appear that that rate should be constant.

It is not constant, for the specific case of constant acceleration.

Why is it that a fast moving body requires more power to accelerate than a still body?

This is more a matter of fact: the same force, acting on increasing speed, correspond to more power. I can try to connect it to another concept (although this does not represent a more explicit answer): the kinetic energy is proportional to $$v^2$$ (square of the speed), thus, as the object moves faster, we need more and more power to give the necessary increase in kinetic energy.

And if we would like to know how much power we need to accelerate that body, should we then use the maximum speed of 2 m/s?

We need an increasing power from the beginning to the moment at which the speed is 2 m/s. So there is not a single value of power. We can evaluate the total energy needed to accelerate to $$v=$$2 m/s: it equals the final kinetic energy, i.e. $$E=1/2 m v^2$$ (where $$m$$ is the mass of the object and $$v$$ the final speed). Dividing this value by the time required for the acceleration, i.e. by $$\Delta t=v/a$$ (where $$a$$ is the acceleration), we get the average power $$E/\Delta t$$.

$$\frac{dv}{dE} = \frac{d}{dE} [(2E/m)^{\frac 1 2}]$$ $$\frac{dv}{dE} = \sqrt{\frac 2 {mE}}$$

which diverges at $$E=0$$ (i.e. $$v=0$$), so the velocity is infinitely sensitive to energy for a body at rest.

You can understand it if you think that the power is the rate of doing work on the accelerated object. This is the rate in time. As the speed increases the distance traveled in the same time interval (1 second for example) increases. But the work done is force times displacement (or distance). So in the same time interval the distance traveled is larger so the work done is higher so the power is higher.

Bear in mind that power as a function of time is the rate at which work is done as a function of time, or

$$P(t)=\frac{dW(t)}{dt}$$

For a constant force $$F$$ acting through a distance $$d$$ the work done $$W$$ is

$$W=Fd$$

If the force is constant, so is the acceleration. The distance $$d$$ a mass travels at constant acceleration given zero initial velocity is given by

$$d=\frac{at^2}{2}$$

That means the work done as a function of time is

$$W(t)=F\frac{at^2}{2}$$

Then power as a function of time, $$P(t)$$ which we gave at the beginning, is given by

$$P(t)=\frac{dW(t)}{dt}=Fat$$

This tells us for a constant force$$F$$, at $$t=0$$ with zero initial velocity, the power is zero and increases linearly with time thereafter.

Hope this helps.

This is a really good question!

Let me phrase it in a slightly different way, because it's the same problem according to calculus. The problem is, if I let go of a ball at rest, then the velocity is zero, so surely it's not moving, so why does it ever move? We would say “yeah, its velocity is zero for an instant, but it is increasing because it is in a state of acceleration.”

Similarly you might say: as we know, the speed is related to the kinetic energy and the change in kinetic energy over time is due to power exerted on that object, but it is at rest: nothing can exert any power on it; so why does it ever move? And the answer is the same: the power exerted is zero for an instant, but it is increasing because it is in a state of acceleration.

In other words when we are looking at a curve at a point, we have a tangent line which is very important, but there are also deviations from those tangent lines which are also important: they are just more important over longer timescales whereas the tangent line is only important over the shortest timescales. We speak of the "order" of the zeros, the ball at rest has a position which is constant to first order in time, but then to second order in time it is not constant. This means that its position is $$y = y_0 + \frac12 a~(t - t_0)^2$$ for some constants $$y_0, t_0, a$$.

Mathematics actually furnishes us with some even weirder examples. For example you might want a curve $$y(x)$$ which is constant to all orders in some variable $$x$$, but is not the constant curve $$y(x) = C$$. Mathematics says that yeah, that's certainly doable, and a great example is the function $$y(x) = \begin{cases} e^{-1/x^2}& x\ne 0\\ 0& x = 0 \end{cases}$$at the point $$x=0$$. Every derivative of this thing at that point is well-defined and zero, but the function still manages to “crawl out of the abyss.”

This decomposition of a function into its behavior over various “orders” in time or space is called a Taylor series, and each Taylor series only approximates a function within a certain “radius of convergence,” which in the above case happens to be 0. If you are doing physics then you generally assume (until you get into stochastic mathematics—in other words, until you start explicitly modeling noise) that the world is perfect and smooth and every function that you're using to model the world has some Taylor series that is valid for a nice big interval.

So that is why without thinking about it we expand to first order, second order, third order... so your teachers have been surreptitiously getting you to think in terms of first-order changes. But to your question, what do we do when the first-order change is zero?, the answer is always going to be well then we need to know more about the second order change. What if that's zero, too? Well, then the third order. And so on.

In you context you would have an average power of $$P_{avg} =F\Delta v$$ where $$\Delta v$$ is the change in velocity. This way, if you accelerate at a constant force $$P_{avg} =F \times 2m/s$$. With no other information about the force or the time you want to spend accelerating the object you can't get a value for $$P_{avg}$$.

• The equation is OK. And the interpretation too. The power varies with velocity so it will increase if the force is constant. $F \Delta v$ is the average power, or $\frac{\Delta E}{\Delta t}$ for the entire interval. – nasu May 29 '19 at 15:44
• The equation is $P=F v$, where $v$ is the speed, not the "change". This is connected to the non-linear increase of kinetic energy, $1/2 m v^2$. – Doriano Brogioli May 29 '19 at 15:48
• OK I'm sorry, I'll edit that... :/ apparently I was the one that misunderstood the question – Ballanzor May 29 '19 at 15:53